Extended Polar Zonohedra

Polar zonohedra have an n-fold
symmetry axis and rhombic faces. There are n faces meeting at the top and
bottom vertex, and n-1 girdles of faces. The total number of faces is thus
n(n-1).

It can be shown that the vertices all lie on equally spaced planes
normal to the polar axis, and that the longitudinal cross section of the
zonohedron is approximately a sine curve.

There are only n distinct edge directions in a polar zonohedron. We can
remove sets of parallel edges and diminish the zonohedron (left) or add a
new set of parallel edges distinct from the original edges and augment
the zonohedron. For the most part the results are of no great interest.

However, there is one exception. If n is odd, we can cut the
zigzag series of edges that bisects the zonohedron, pull the two halves
apart parallel to the axis, and add a set of edges joining the two halves.
If the new edges are the same length as the original edges, the solid
still is equilateral. We add a girdle of 2n faces around the equator with
its zone axis parallel to the polar axis.

For a simple polar zonohedron with n-fold symmetry there are two independent
dimensions: the radius R and the polar axis length L. We define angle A = 360/n.
We can find angle Q, the angle between one of the top or bottom edges and the
polar axis. We can then find Vi, the face angle of the i-th girdle of rhombuses,
and Ri, the radius of the i-th ring of vertices. The formulas are defined
for edge length = 1.

Now, what are the face angles for the new girdle? We can plow through some
geometry or we can think a bit (oww - make the pain go away!). The new
girdle rhombuses consist of one of the original edges plus a new edge parallel
to the polar axis. All the original edges make an angle Q with the polar axis.
So the face angles of the equatorial girdle must be Q and 180-Q. That was a lot
easier than calculating.

The most interesting extended zonohedron is the n=5 case. If
we solve for the case where the faces are congruent, we get, not the
rhombic triacontahedron, but a flattened solid with 20 faces

For n=5, we have: A=72, A/2=36, Sin(A/2)=0.5877....

For simplicity, let sin36=v and sin72=w

For the isohedral case, solve for V1=180-V2

SinV1/2=Rsin(A/2)sin(A/2)=Rv2

SinV2/2=Rsin(A/2)sin(2A/2)=Rvw=(w/v)SinV1/2

Sin(V1/2)=sin((180-V2)/2)=sin(90-V2/2)=cos(V2/2)

SinV2/2=(w/v)sin(V1/2), so sin2(v2/2)=(w/v)2sin2(v1/2)

cos2(v2/2)=1-(w/v)2sin2(v1/2)=sin2(v1/2)

Thus sin2(v1/2)=1/(1+(w/v)2), and sin(v1/2)=0.5257

V1/2=31.72, thus V1=63.43, the angles for a rhombic triacontahedron.

Now let's calculate R and L* for the extended case.

Sin(V1/2) = sinQsin(A/2), so sinQ=Sin(V1/2)sin(A/2)=.8944

Q = 63.43 degrees also. Thus the equatorial girdle faces are congruent
with all the other faces.

Cos Q = 0.4772

L* = 5cosQ+1 = 3.236 = sqrt(5)+1

R = sinQ/sin(A/2) = 0.8944/sin(36) = 1.5217

It turns out that we get the rhombic triacontahedron by
extending the 20-sided simple zonohedron.

This sort of thing only works for n odd. What happens if we try it with a
rhombohedron (n=3)?

The rhombohedron has six faces. We pull the solid apart and add a girdle of
six more around the middle for a total of n=12. We can get a solid with
congruent faces, and it turns out to be the rhombic dodecahedron viewed along
one of its threefold symmetry axes.

So the rhombic dodecahedron is both a simple 4-zonohedron and an extended
3-zonohedron.

For n=3, we have: A=120, A/2=60, Sin(A/2)=sqrt(3)/2. We only have one girdle
of faces, so we have to find the case V1=Q.

V1=Q

Sin(V1/2)=Rsin(A/2)sin(A/2)=3R/4

sinQ = Rsin(A/2)

But we can also write cosQ=cos(V1)=cos2(V1/2)

cosQ=cos2(V1/2)=1-2sin2(V1/2)=1-2(9R2/16)=1-9R2/8

Also, since sinQ = Rsin(A/2), cos2Q=1-3R2/4

Thus (3/2)cos2Q=1/2+cosQ or 3cos2Q-2cosQ-1=0

This factors to (3cosQ+1)(cosQ-1)

Solving, we get cosQ=1 or 1/3. -CosQ=1 means Q=0, cosQ=1/3 means
Q=109.471, the obtuse face angle of a rhombic dodecahedron. We get the
obtuse angle because that's the one at the polar vertex.